Find $\lim_{h\to 0}\dfrac{3\ln(e+h)-3\ln(e)}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac1e$ (Choice B) B $\dfrac3e$ (Choice C) C $e$ (Choice D) D The limit doesn't exist
Explanation: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $3\ln(e+h)-3\ln(e)$, we can tell that the function is $f(x)=3\ln(x)$ and the $x$ -value is $e$. In other words, the limit expression is equal to $f'(e)$ for $f(x)=3\ln(x)$. Let's find $f'(x)$ : $f'(x)=3\cdot \dfrac1x=\dfrac3x$ Now let's evaluate $f'(e)$ : $f'(e)=\dfrac3e$ In conclusion, $\lim_{h\to 0}\dfrac{3\ln(e+h)-3\ln(e)}{h}=\dfrac3e$.